Author Topic: Integration Questions. (Read 2319 times) Tweet Share

gta007 · « **on:** May 31, 2008, 12:38:49 pm »

Guys/gals been stuck on these for quite a while. Any help is appreciated.

1. Using:
$\[ n\int \cos^n(x) dx = sin(x)cos^{n-1}(x) +(n-1)\int \cos^{n-2}(x) dx$

Evaluate. This is part (iv) of the question:
$\int_0^{\pi/4} sec^4(x) \,dx$

Unsure of what to do, do I change $sec^4(x)$ to $cos^{-4}(x)$ ?

2. (a) I have simplified $\frac{1}{1+tan(x)}}$ + $\frac{1}{1+cot(x)}}$ $= 1$

Part (b) asks: Let $\phi = \pi/2 - \theta$ Show that $\[ \int_0^{\pi/2} \frac{d\theta}{1+tan\theta} =$ $\int_0^{\pi/2} \frac{d\phi}{1+cot\phi}$

(c) Use these results to evaluate $\[ \int_0^{\pi/2} \frac{d\theta}{1+tan\theta}$

Gosh, latex is hard to use.

Mao · « **Reply #1 on:** May 31, 2008, 01:09:26 pm »

1)
i dont see how the relationship will help this question, a simple trig identity is enough:

$\begin{align*} \int sec^4(x)\ dx &= \int sec^2(x) \cdot (tan^2(x)+1)\ dx \mbox{, and substituting }u=\tan(x) \\ &= \int u^2 + 1 \ du \\ &= \frac{\tan^3(x)}{3} + \tan(x) + c \\ \int_0^{\pi/4} sec^4(x)\ dx &= \left[ \frac{\tan^3(x)}{3} + \tan(x) \right]_0^{\pi/4} \\ &= \frac{4}{3}\\ \end{align*} $

2)
knowing that $cot(x) = \tan\left( \frac{\pi}{2}-x\right)$

b):
$\begin{align*} RHS &= \int_0^{\pi/2} \frac{d\phi}{1+cot \phi}\mbox{, substituting }\phi = \frac{\pi}{2}-\theta \\ &= \int_{\pi/2-0}^{\pi/2-\pi/2} \frac{d\phi}{1+\tan\left(\frac{\pi}{2}-\left(\frac{\pi}{2}-\theta\right) \right)} \cdot \frac{d\theta}{d\phi} \\ &= \int_{\pi/2}^0 \frac{d\theta}{1+\tan(\theta)} \cdot -1 \\ &= \int_{0}^{\pi/2} \frac{d\theta}{1+\tan(\theta)}\\ \end{align*} $

c):
using the relationship from part 1:

$\begin{align*} \int_{0}^{\pi/2} \frac{d\theta}{1+\tan(\theta)} + \int_{0}^{\pi/2} \frac{d\theta}{1+cot(\theta)} &= \int_{0}^{\pi/2} d\theta \\ \int_{0}^{\pi/2} \frac{d\phi}{1+cot(\phi)} + \int_{0}^{\pi/2} \frac{d\theta}{1+cot(\theta)} &= \frac{\pi}{2} \\ 2\int_{0}^{\pi/2} \frac{d\phi}{1+cot(\phi)} &= \frac{\pi}{2} \\ \int_{0}^{\pi/2} \frac{d\phi}{1+cot(\phi)} &= \frac{\pi}{4} \\ \implies \int_{0}^{\pi/2} \frac{d\theta}{1+\tan(\theta)} &= \frac{\pi}{4} \\ \end{align*} $
since both integrals [in step two] can be considered as "area" functions, the difference between $\phi$ and $\theta$ does not play a part

cara.mel · « **Reply #2 on:** May 31, 2008, 01:30:28 pm »

1. Going along with your idea of $cos^{-4}(x)$ :
(btw this is my first real attempt at latex as well, not my fault the maths people didn't beat me for once. People, why do my dx etc seem so close to everything else)
(btw, I've either done the same question, or something very close to it, in an assignment this year(with the help of Neobeo

))

$n\int \cos^{n}(x) dx = sin(x)cos^{n-1}(x) + (n-1) \int \cos^{n-2}(x) dx$
=> $(n-1) \int \cos^{n-2}(x) dx = sin(x)cos^{n-1}(x) -n\int \cos^{n}(x) dx$
=> $\int \cos^{n-2}(x) dx = \frac{sin(x)cos^{n-1}(x) -n\int \cos^{n}(x) dx}{(n-1)}$

Now putting the numbers in:
$\int \cos^{-4}(x) dx = \frac{sin(x)cos^{-3}(x) + 2 \int \cos^{-2}(x) dx}{(-3)}$
This makes $\int \cos^{-2}(x) dx$ , so we go find that out:
$\int \cos^{-2}(x) dx = \frac{sin(x)cos^{-1}(x) + 0 \int \cos^{-0}(x) dx}{(-1)}$
$\int \cos^{-2}(x) dx = -{sin(x)cos^{-1}(x)$

We can now put the value of $\int \cos^{-2}(x) dx$ back into our original one, so that
$\int \cos^{-4}(x) dx = \frac{sin(x)cos^{-3}(x) + 2 \int \cos^{-2}(x) dx}{(-3)} = \frac{sin(x)cos^{-3}(x) -2{sin(x)cos^{-1}(x)}}{(-3)}$
So then
$\int_{0}^{pi/4} cos^-4(x) = \left[ \frac{sin(x)cos^{-3}(x) -2{sin(x)cos^{-1}(x)}}{(-3)} \right]_0^{\pi/4}$ (I give up on latex here, I can't see it in the guide, but you kinda get what I'm trying to write?

)
And then you sub in numbers ^_^

mod edit: fixed

and added to guide =]

gta007 · « **Reply #3 on:** May 31, 2008, 01:55:12 pm »

Thanks for the help guys.
Yeh, in Q1 i know that sec^4(x) can be broken down into sec^2(x)(tan^2(x) +1). But the stupid book wanted to do it this way. (Pg 281 Essential).

One question, in 2(c) on the second line, how did you get it to equal pi/2?

caramel, thanks for the help. I didn't go into it that far. Will look into it now.

Thank you all.

cara.mel · « **Reply #4 on:** May 31, 2008, 02:01:09 pm »

I'm glad you understand mao's one because I have no idea, it has magical u's from thin air in it

Mao · « **Reply #5 on:** May 31, 2008, 02:01:49 pm »

Quote from: gta007 on May 31, 2008, 01:55:12 pm

One question, in 2(c) on the second line, how did you get it to equal pi/2?

$\int_0^{\pi/2} d\theta = \left[\theta\right]_0^{\pi/2} = \frac{\pi}{2}$

cara.mel · « **Reply #6 on:** May 31, 2008, 02:03:43 pm »

Quote from: caramel on May 31, 2008, 02:01:09 pm

I'm glad you understand mao's one because I have no idea, it has magical u's from thin air in it

Oh I get it now, you have like 3 steps in one line

gta007 · « **Reply #7 on:** May 31, 2008, 02:18:57 pm »

Just wondering, how does

$\int_{0}^{\pi/2} \frac{d\phi}{1+cot(\phi)}$ = $\int_{0}^{\pi/2} \frac{d\theta}{1+cot(\theta)}$

gta007 · « **Reply #8 on:** May 31, 2008, 03:08:59 pm »

caramel, your way worked out.

Just some slight errors in the working out, but it helped me work my way to the answer.
In step you it should have been $(n-1) \int \cos^{n-2}(x) dx = n\int \cos^{n}(x) - sin(x)cos^{n-1}(x) dx$

For those interested the final step should be:

$\int_{0}^{pi/4} cos^-4(x) = \left[ \frac{-sin(x)cos^{-3}(x) -2{sin(x)cos^{-1}(x)}}{(-3)} \right]_0^{\pi/4}$

dcc · « **Reply #9 on:** May 31, 2008, 03:53:01 pm »

$\begin{align*} -4 \int \cos^{-4} x\mbox{ } dx &= \sin x \cos^{-5} x - 5 \int \cos^{-6} x\mbox{ }dx\\ &= \sin x \cos^{-5} x + \dfrac{5}{6} \int -6 \cos^{-6} x\mbox{ }dx\\ &= \sin x \cos^{-5} x + \dfrac{5}{6} \left(\sin x \cos^{-7} x - 7 \int \cos ^{-8} x\mbox{ }dx\right)\\ &= \sin x \cos^{-5} x + \dfrac{5}{6} \sin x \cos^{-7} x + \dfrac{35}{48} \int -8 cos^{-8}x\mbox{ }dx\\ &= \sin x \cos^{-5} x + \dfrac{5}{6} \sin x \cos^{-7} x + \dfrac{35}{48} \left(\sin x \cos^{-9} x - 9 \int \cos^{-10} x\mbox{ }dx\right)\\ &= \sin x \cos^{-5} x + \dfrac{5}{6} \sin x \cos^{-7} x + \dfrac{35}{48} \sin x \cos^{-9} x - \dfrac{105}{16} \int \cos^{-10} x\mbox{ }dx\\ &= \sin x \sum_{k=0}^{m} \left[\left(\dfrac{8 \cdot \Gamma \left(\dfrac{5}{2} + k \right)}{3\sqrt{\pi} \cdot \Gamma \left(3 + k \right)} \right) \cos^{-(5 + 2k)} x\right] - (5 + 2m) \cdot \left(\dfrac{8 \cdot \Gamma \left(\dfrac{5}{2} + m \right)}{3\sqrt{\pi} \cdot \Gamma \left(3 + m \right)} \right) \int \cos^{-(6+2m)} x\mbox{ }dx\right]\\ \end{align*}$

Where m >= 0 and an integer.

I was going to go somewhere from this, but the techniques required elude me. I will edit this in the future and continue.

Collin Li · « **Reply #10 on:** May 31, 2008, 03:55:39 pm »

Quote from: gta007 on May 31, 2008, 02:18:57 pm

Just wondering, how does

$\int_{0}^{\pi/2} \frac{d\phi}{1+cot(\phi)}$ = $\int_{0}^{\pi/2} \frac{d\theta}{1+cot(\theta)}$

Let $\phi = \theta$ ? It makes no difference whether I talk about an integral in terms of $x$ , $u$ , $\theta$ or

(meow).

Mao · « **Reply #11 on:** May 31, 2008, 06:26:06 pm »

Quote from: gta007 on May 31, 2008, 02:18:57 pm

Just wondering, how does

$\int_{0}^{\pi/2} \frac{d\phi}{1+cot(\phi)}$ = $\int_{0}^{\pi/2} \frac{d\theta}{1+cot(\theta)}$

this is only true as we are evaluating definite integrals.

by the second fundamental theorem of calculus:

$\int_a^b f(x)\ dx = F(b)-F(a)$

that is, if we used $t$ instead of $x$ , it'll have no bearing on the result, as

$\int_a^b f(t)\ dt = F(b)-F(a)$

that's the basic principle.

[and I think there are limitations to this, but i dont know exactly what they are. it's not the most pedantic thing to do, but recognising that for our "simple" applications in spec. math it works, should be enough.]

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